# Intersection pointsFind the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x - 2

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You need to check the relative position between the line and parabola, hence, you need to solve the system of equations, such that:

`{(y = x^2 - 5x + 4),(y = 2x - 2):}`

`x^2 - 5x + 4 = 2x - 2 => x^2 - 5x - 2x + 4 + 2 = 0`

`x^2 - 7x + 6 = 0`

You may use quadratic formula to evaluate the solutions to quadratic equation, such that:

`x_(1,2) = (7 +- sqrt(49 - 24))/2 => x_(1,2) = (7 +- sqrt25)/2`

`x_(1,2) = (7 +-5)/2 => x_1 = 6 ; x_2 = 1`

You need to find the y coordinates such that:

`x_1 = 6 => y_1 = 2x_1 - 2 => y_1 = 12 - 2 => y_1 = 10`

`x_2 = 1 => y_2 = 2x_2 - 2 => y_2 = 2 - 2 = 0`

**Hence, the line `y = 2x - 2` is secant to parabola `y = x^2 - 5x + 4` because it intersects parabola at `(1,0)` and `(6,10)` .**

The common point that lies on the line and parabola in the same time is the intercepting point of the line and parabola.

So, the y coordinate of the point verify the equation of the line and the equatin of the parabola, in the same time.

2x-2 = x^2-5x+4

We'll factorize by 2 both sides:

2(x-1) = x^2-5x+4

But the roots of the quadrstic are x1 = 1 and x2 = 4.

2(x-1) =(x-1)(x-4)

We'll divide by (x-1):

2 = x - 4

We'll subtract 2 both sides and we'll use the symmetric property:

x - 6 = 0

x = 6

Now, we'll substitute in the equation of the line to determine y:

y = 2x-2

y = 2*6 - 2

y = 10

The common point between the line and parabola is (6 , 10).